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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3o4259

The alcohol shown is dehydrated with acid, and hydride shift is allowed. Counting E and Z isomers as different compounds, select all correct statements.

  1. AWithout rearrangement three alkenes can form.
  2. BIncluding alkenes from the hydride-shifted cation, four alkenes can form in total.
  3. C2-Methylpent-1-ene can form only after rearrangement.
  4. D2-Methylpent-2-ene exists as separate E and Z isomers.

Answer: A, B, C

The alcohol is 2-methylpentan-3-ol. The unrearranged cation gives 2-methylpent-2-ene and (E)- and (Z)-4-methylpent-2-ene, three alkenes. A hydride shift gives the tertiary 2-methylpent-2-yl cation, which adds 2-methylpent-1-ene, for four in total. 2-Methylpent-2-ene has two identical methyl groups on one carbon, so it has no E/Z isomers.
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